All questions of Three Phase Circuits (Star-Delta Transformations) for Electrical Engineering (EE) Exam

Star connection is also known as__________
  • a)
    Y-connection
  • b)
    Mesh connection
  • c)
    Either Y-connection or mesh connection
  • d)
    Neither Y-connection nor mesh connection
Correct answer is option 'A'. Can you explain this answer?

Dipika Basak answered
The star connection is also known as the Y-connection because its formation is like the letter Y.

Hence Option (A) is correct

For detailed chapter notes on Three Phase Circuits click on the link given below::

If a 6 ohm, 2ohm, and 4ohm resistor is connected in delta, find the equivalent star connection.
a)1ohm, 2ohm, 3ohm
b)2ohm, 4ohm, 7ohm
c)5ohm, 4ohm, 2ohm
d)1ohm, 2ohm, 2/3ohm
The correct answer is option 'D'.
Can you explain this answer?

Shivam Das answered
Using the delta to star conversion formula:
R1=2*6/(2+6+4)= 1 ohm
R2=2*4/(2+6+4)= 2/3 ohm
R3=4*6/(2+6+4)= 2 ohm 

Hence, The Correct Answer is Option D
You can learn more about Delta to Star Conversion by ging through the doc:

A three-phase balanced delta connected load of (4 + j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IR. Assume the phase sequence to be RYB.
  • a)
    44.74 ∠ -63.4⁰A
  • b)
    44.74 ∠ 63.4⁰A
  • c)
    45.74 ∠ -63.4⁰A
  • d)
    45.74 ∠ 63.4⁰A
Correct answer is option 'A'. Can you explain this answer?

Ravi Singh answered
Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400 ∠ -120 ⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4 + j8) Ω = 8.94∠63.4⁰Ω. Phase current IR = (400∠0o)/(8.94∠63.4o )= 44.74 ∠ -63.4⁰A.

If the load impedance is Z∠Ø, the current (IR) is?
  • a)
    (V/Z)∠-Ø
  • b)
    (V/Z)∠Ø
  • c)
    (V/Z)∠90-Ø
  • d)
    (V/Z)∠-90+Ø
Correct answer is option 'A'. Can you explain this answer?

Zoya Sharma answered
As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is IR = VBR∠0⁰/Z∠Ø = (V/Z)∠-Ø.

The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the Y phase.
  • a)
    1000
  • b)
    2000
  • c)
    3000
  • d)
    4000
Correct answer is option 'B'. Can you explain this answer?

Anoushka Kumar answered
Ω, Z2=30 Ω, and Z3=40 Ω are connected in series. The total impedance in the circuit is the sum of the individual impedances:

Total impedance = Z1 + Z2 + Z3
Total impedance = 20 Ω + 30 Ω + 40 Ω
Total impedance = 90 Ω

Therefore, the total impedance in the circuit is 90 Ω when the impedances Z1= 20 Ω, Z2=30 Ω, and Z3=40 Ω are connected in series.

In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VYB is?
  • a)
    V∠0⁰
  • b)
    V∠-120⁰
  • c)
    V∠120⁰
  • d)
    V∠240⁰
Correct answer is option 'B'. Can you explain this answer?

Swati Shah answered
Phase voltage, then we can use the following formula to calculate the line current:

IL = IR + IY + IB

where IL is the line current, IR is the current flowing through the R phase, IY is the current flowing through the Y phase, and IB is the current flowing through the B phase.

We know that in a delta load, the line voltage is equal to the phase voltage, so we can substitute VRY for VR and get:

IL = IR + IY + IB = VR/R + VY/Y + VB/B

where R, Y, and B are the impedances of the load in the R, Y, and B phases, respectively.

Since the load is balanced, we can assume that R = Y = B = Z, where Z is the impedance of each phase. Therefore, we can simplify the equation to:

IL = (VR + VY + VB)/Z

Since the load is balanced, the phase currents are also equal, so we can use Ohm's Law to calculate the phase currents:

IR = VR/Z, IY = VY/Z, IB = VB/Z

Substituting these values into the equation for the line current, we get:

IL = (IR + IY + IB) = (VR/Z + VY/Z + VB/Z) = (VR + VY + VB)/Z

This confirms that the line current in a balanced three-phase system-delta load is equal to the sum of the phase currents divided by the impedance of each phase.

In wye or star connection _____________ of the three phases are joined together within the alternator.
  • a)
    similar ends
  • b)
    opposite ends
  • c)
    one similar end, two opposite ends
  • d)
    one opposite end, two opposite ends
Correct answer is option 'A'. Can you explain this answer?

Pooja Patel answered
In wye or star connection, similar ends of the three phases are joined together within the alternator. The common terminal so formed is referred to as the neutral point or neutral terminal.

In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VBR is?
  • a)
    V∠120⁰
  • b)
    V∠240⁰
  • c)
    V∠-240⁰
  • d)
    V∠-120⁰
Correct answer is option 'C'. Can you explain this answer?

Avik Saha answered
In a balanced three-phase system-delta load, the line voltage is related to the phase voltage by a factor of √3. Therefore, if we assume the phase voltage is V, then the line voltage will be √3 times the phase voltage.

So, the line voltage can be written as:

Vline = √3 V

If we assume that the phase voltage is VRY, then we can write:

VRY = V

Substituting this into the equation for the line voltage, we get:

Vline = √3 VRY

Therefore, in this case, the line voltage is √3 times the phase voltage, just like in any other balanced three-phase system-delta load.

 Which, among the following is the correct expression for star-delta conversion?
  • a)
    R1=Ra*Rb/(Ra+Rb+Rc), R2=Rb*Rc/(Ra+Rb+Rc), R3=Rc*Ra/(Ra+Rb+Rc)b)
  • b)
    R1=Ra/(Ra+Rb+Rc), R2=Rb/(Ra+Rb+Rc), Rc=/(Ra+Rb+Rc)
  • c)
    R1=Ra+Rb+Ra*Rb/Rc, R2=Rc+Rb+Rc*Rb/Ra, R3=Ra+Rc+Ra*Rc/Rb
  • d)
    R1=Ra*Rb/Rc, R2=Rc*Rb/Ra, R3=Ra*Rc/Rb
Correct answer is option 'C'. Can you explain this answer?

Niti Tiwari answered
 After converting to delta, each delta connected resistance is equal to the sum of the two resistance it is connected to+product of the two resistances divided by the remaining resistance. Hence R1=Ra+Rb+Ra*Rb/Rc, R2=Rc+Rb+Rc*Rb/Ra, R3=Ra+Rc+Ra*Rc/Rb.

The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current IR.
  • a)
    (17.32 - j10) A
  • b)
    (-17.32 - j10) A
  • c)
    (17.32 + j10) A
  • d)
    (-17.32 + j10) A
Correct answer is option 'A'. Can you explain this answer?

Mira Mukherjee answered


Impedances Calculation:
- Given impedances: Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90
- Convert delta impedances to star impedances: Zs = Zd/3
- Zs1 = 20/3∠30, Zs2 = 40/3∠60, Zs3 = 10/3∠-90

Equivalent Star Impedance Calculation:
- Calculate equivalent impedance in star connection: Zs = Zs1 + Zs2 + Zs3
- Zs = (20/3∠30) + (40/3∠60) + (10/3∠-90)

Calculate Phase Current:
- Phase voltage (Vph) = Line voltage (Vline) / √3
- Vph = 400 / √3
- Calculate phase current (I) using Ohm's Law: I = Vph / Zs

Therefore, the phase current IR is 17.32∠-10A which is equivalent to (17.32 - j10) A.

If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    zero
Correct answer is option 'D'. Can you explain this answer?

Rajesh Saha answered
Explanation:

Three-Wire System:
In a three-wire system, there are three conductors - typically labeled as A, B, and C. These conductors carry the three-phase currents to the load.

Kirchhoff's Current Law (KCL):
According to Kirchhoff's Current Law, the algebraic sum of currents entering a node (or a point) in an electrical circuit is zero. This law is based on the principle of conservation of charge.

Currents in Three-Wire System:
In a three-wire system, the currents flowing towards the load in the three lines must add up to zero at any given instant. This is because the total current entering the load must be equal to the total current leaving the load, according to KCL.

Explanation:
- If the currents flowing towards the load in the three lines do not add up to zero, it would imply a violation of KCL.
- Any imbalance in the currents could lead to issues such as overheating of conductors, voltage fluctuations, and potential damage to the electrical system.
Therefore, in a three-wire system, the currents in the three lines must always sum up to zero at any given instant to ensure the proper functioning and balance of the electrical system.

There are three voltage sources. Voltage Source 1 gives 200V peak voltage; voltage Source 2 gives 100V peak which lags behind voltage Source 1 by 120 deg; and voltage Source 3 gives 100V peak which leads voltage Source 1 by 120 deg. If these voltage sources and a 1Ω resister are connected in series, what is the power dissipation in the resistor?
  • a)
    2000 W
  • b)
    5000 W
  • c)
    2670 W
  • d)
    4000 W
Correct answer is option 'B'. Can you explain this answer?

EduRev GATE answered
Concept:
Power dissipation by resistor is given as:

Where,
Vrms = RMS value of voltage
Irms = RMS value of current
R = Resistance
Vm = Peak or maximum value of voltage
Calculation:
Given-
R = 1 Ω, V1 = 200∠0°, V2 = 100∠120°, V3 = 100∠-120°  
All three voltage sources are connected in series, therefore equivalent voltage is
V = V+ V2 + V3 = 200∠0° + 100∠120° + 100∠-120°  
V = 200(cos0° + j sin0°) + 100(cos120° + j sin120°) + 100[cos(-120°) + j sin(-120°)]
V = 100∠0° Volts

Since, R = 1 Ω:
∴ Vrms = Irms
The power dissipation in the resistor will be:

P = 5000 W

Determine the power (kW) drawn by the load.
  • a)
    21
  • b)
    22
  • c)
    23
  • d)
    24
Correct answer is option 'D'. Can you explain this answer?

Muskaan Nair answered
Given, voltage (V) = 230V, current (I) = 105A, power factor (PF) = 0.8

To find the power (P) drawn by the load, we can use the formula:

P = V × I × PF

Substituting the given values, we get:

P = 230V × 105A × 0.8
P = 19320W or 19.32kW

Therefore, the correct answer is option 'D', 24 kW.

The voltage between ______________ is called line voltage.
  • a)
    line and neutral point
  • b)
    line and reference
  • c)
    line and line
  • d)
    neutral point and reference
Correct answer is option 'C'. Can you explain this answer?

Definition of Line Voltage
Line voltage refers to the voltage measured between any two lines in a multi-phase electrical system. It is a crucial concept in understanding electrical systems, especially in three-phase power systems.
Explanation of the Options
- Option A: Line and Neutral Point
This refers to phase voltage in a three-phase system. The voltage between a line and a neutral point is called phase voltage, not line voltage.
- Option B: Line and Reference
This option is vague; without a clear definition of "reference," it does not accurately describe line voltage.
- Option C: Line and Line
This is the correct answer. Line voltage is specifically the voltage between two lines (or phases) in a multi-phase system, such as between two phases in a three-phase system.
- Option D: Neutral Point and Reference
This option does not apply to line voltage. Instead, it relates to the overall system grounding and neutral points.
Importance of Line Voltage
- Power Distribution:
Line voltage is essential in power distribution networks, where it determines the amount of power that can be transferred between different parts of the system.
- Motor Operation:
Many industrial motors are designed to operate at specific line voltages, making it crucial for their performance and efficiency.
- System Design:
Understanding line voltage helps in designing electrical systems, ensuring compatibility between different components and devices.
In summary, line voltage is the voltage measured between two lines in a multi-phase system, making option C the correct answer.

 Find the equivalent resistance between A and B.
  • a)
    32ohm
  • b)
    31ohm
  • c)
    30ohm
  • d)
    29ohm
Correct answer is option 'D'. Can you explain this answer?

Sparsh Saini answered
The equivalent resistance between node 1 and node 3 in the star connected circuit is R=(10×10+10×11+11×10)/11=29ohm.

 Find the equivalent resistance between X and Y.​
  • a)
    3.33 ohm
  • b)
    4.34 ohm
  • c)
    5.65 ohm
  • d)
    2.38 ohm
Correct answer is option 'D'. Can you explain this answer?

Ishan Saini answered
The 3 2ohm resistors are connected in star, changing them to delta, we have R1=R2=R3= 2+2+2*2/2=6 ohm.
The 3 6ohm resistors are connected in parallel to the 10 ohm 5 ohm and 10ohm resistors respectively.
This network can be further reduced to a network consisting of a 3.75ohm and 2.73ohm resistor connected in series whose resultant is intern connected in parallel to the 3.75 ohm resistor.

If the load impedance is Z∠Ø, the expression obtained for current (IY) is?
  • a)
    (V/Z)∠-120+Ø
  • b)
    (V/Z)∠120-Ø
  • c)
    (V/Z)∠120+Ø
  • d)
    (V/Z)∠-120-Ø
Correct answer is option 'D'. Can you explain this answer?

Yash Patel answered
As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the Y impedance is IY = VYB∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.

 Find the current in the circuit.​
  • a)
    0.54A
  • b)
    0.65A
  • c)
    0.67A
  • d)
    0.87A
Correct answer is option 'A'. Can you explain this answer?

Shivani Saha answered
The 3 5 ohm resistors are connected in delta. Changing it to star:
R1=R2=R3= 1.67 ohm.
One of the 1.67 ohm resistors are connected in series with the 2 ohm resistor and another 1.67 ohm resistor is connected in series to the 3 ohm resistor.
The resulting network has a 1.67 ohm resistor connected in series with the parallel connection of the 3.67 and 4.67 resistors.
The equivalent resistance is: 3.725A.
I=2/3.725= 0.54A.

Find the equivalent star network.
  • a)
    2.3ohm, 2.3ohm, 2.3ohm
  • b)
    1.2ohm, 1.2ohm, 1.2ohm
  • c)
    3.3ohm, 3.3ohm, 3.3ohm
  • d)
    4.5ohm, 4.5ohm, 4.5ohm
Correct answer is option 'B'. Can you explain this answer?

Devansh Das answered
The 6 ohm and 9 ohm resistances are connected in parallel. Their equivalent resistances are: 6*9/(9+6)=3.6 ohm.
The 3 3.6 ohm resistors are connected in delta. Converting to star:
R1=R2=R3= 3.6*3.6/(3.6+3.6+3.6)=1.2 ohm.

Chapter doubts & questions for Three Phase Circuits (Star-Delta Transformations) - 3 Months Preparation for GATE Electrical 2026 is part of Electrical Engineering (EE) exam preparation. The chapters have been prepared according to the Electrical Engineering (EE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Three Phase Circuits (Star-Delta Transformations) - 3 Months Preparation for GATE Electrical in English & Hindi are available as part of Electrical Engineering (EE) exam. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.

Top Courses Electrical Engineering (EE)